What will be the product in the following reaction ?

The correct answer is Option (1) → 2,4,6-tribromophenol

The –OH group of phenol releases electrons into the ring by resonance. This makes the ring highly reactive towards electrophilic substitution and directs the incoming group to the ortho and para positions. How far the bromination goes depends on the solvent:

In bromine water, water polarises the bromine molecule and partly ionises phenol to the even more reactive phenoxide ion, so all three ortho and para positions are substituted; a solvent of low polarity at low temperature stops the reaction at monobromination.

    \( \displaystyle \ce{C6H5OH + 3Br2} \xrightarrow{\ce{H2O}} \ce{C6H2Br3OH v + 3HBr} \)

Here the reagent is \( \ce{Br2} \)/\( \ce{H2O} \), so the product A is 2,4,6-tribromophenol.

Why the other options are incorrect

• Option (2) ortho-bromophenol — Incorrect. It is formed only when bromination is carried out in a solvent of low polarity, and even then it is the minor product.

• Option (3) para-bromophenol — Incorrect. It is the major product in \( \ce{CHCl3} \) or \( \ce{CS2} \) at low temperature, not in bromine water.

• Option (4) both ortho-bromophenol & para-bromophenol — Incorrect. This mixture is obtained with \( \ce{Br2} \) in a solvent of low polarity. In bromine water the reaction does not stop at one bromine atom.

NCERT (2026-27 edition):
Halogenation